Answer:
The sum of the first 499 terms is 249001
Step-by-step explanation:
* Lets revise the arithmetic sequence
- There is a constant difference between each two consecutive
numbers
- Ex:
# 2 , 5 , 8 , 11 , ……………………….
# 5 , 10 , 15 , 20 , …………………………
# 12 , 10 , 8 , 6 , ……………………………
* General term (nth term) of an Arithmetic sequence:
- U1 = a , U2 = a + d , U3 = a + 2d , U4 = a + 3d , U5 = a + 4d
- Un = a + (n – 1)d, where a is the first term , d is the difference
between each two consecutive terms n is the position of the
number
- The sum of first n terms of an Arithmetic sequence is calculate from
Sn = n/2[a + l], where a is the first term and l is the last term
* Now lets solve the problem
∵ The terms of the sequence are 1 , 3 , 5 , 7 , ......... , 997
∵ The first term is 1 and the second term is 3
∴ The common difference d = 3 - 1 = 2
∵ The first term is 1
∵ The last term is 997
∵ The common difference is 2
- Lets find how many terms in the sequence
∵ an = a + (n - 1) d
∴ 997 = 1 + (n - 1) 2 ⇒ subtract 1 from both sides
∴ 996 = (n - 1) 2 ⇒ divide both sides by 2
∴ 498 = n - 1 ⇒ add 1 for both sides
∴ n = 499
∴ The sequence has 499 terms
- Lets find the sum of the first 499 terms
∵ Sn = n/2[a + l]
∵ n = 499 , a = 1 , l = 997
∴ S499 = 499/2[1 + 997] = 499/2 × 998 = 249001
* The sum of the first 499 terms is 249001
